In triangle $ABC,$ $AP = BQ = \frac{PQ}{2}$ and $CR/RA = 3/2.$ Find the ratio of the area of triangle $BQC$ to the area of triangle $CRQ$
[asy]
pair A, X, Y, B, Z, C;
A = (0,0);
X = (0.3,-0.2);
Z = (0.4,0.2);
Y = 3*X;
B = 4*X;
C = 2.5*Z;
draw(X--Z--Y--C--B--A--C);
label("$A$",A,W);
label("$B$",B,S);
label("$C$",C,N);
label("$P$",X,SW);
label("$Q$",Y,SW);
label("$R$",Z,NW);
[/asy]
Please Try to Respond as Fast as possible.
+2864 In triangle \(ABC\), \(AP = BQ = \frac{PQ}{2}\) and \(CR/RA = 3/2 \). Find the ratio of the area of triangle \(BQC\) to the area of triangle \(CRQ\).
Since AP = BQ = PQ/2. We can conclude that QB is 1/4 of AB.
Let the area of ABC = x
So BQC = \(\frac{x}{4}\)
Area of triangle CAQ = \(x-\frac{x}{4}=\frac{3x}{4}\)
Since CR/RA = 3/2, that means that CRQ = \(\frac{3x}{4}*\frac{3}{5}=\frac{9x}{20}\)
SO
Area of BQC is \(\frac{x}{4}\)
Area of CRQ is \(\frac{9x}{20}\)
Can you find the ratio now? (Hint: It's simple algebra!)
